1 mole of an ideal gas is at 4 atm pressure, occupies 5 L and has an internal en

Question

1 mole of an ideal gas is at 4 atm pressure,
occupies 5 L and has an internal energy of
633.5 J. The gas is first cooled at constant
volume until its pressure is 1 atm. It is then
allowed to expand at constant pressure until
its volume is 7 L with an internal energy of
1465.5 J.
Find the heat added during this process.
Answer in units of J

Explanation / Answer

Change in internal energy equals heat added to the gas minus the work done by it. Delta U = Q - W => Q = Delta U + W Change in internal energy is: Delta U = U_final - U_initial = 1465.5 J - 633.5 J = 832 J Work done by the gas is given the integral W = Delta P dV from V_initial to V_final In the first part of the process the volume is constant, so no work is done. The second part is a constant pressure process. For such a process the work integral simplifies to: W = P * dV = P* (V_final - V_initial) For this process: W = 3*101325Pa *(7

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.